What are the possible numbers of positive, negative, and complex zeros of f(x) = x6 + x5 + x4 + 4x3 − 12x2 + 12? (6 points) positive: 2 or 0; negative: 4, 2, or 0; complex: 6, 4, 2, or 0 positive: 4, 2, or 0; negative: 2 or 0; complex: 4, 2, or 0 positive: 2 or 0; negative: 2 or 0; complex: 4, 2, or 0 positive: 2 or 0; negative: 0; complex: 6, 4, 2, or 0

Answer : C

For f(x), there will be 6 possible zeros

To get positive root we look at the sign changes in coefficients and count the sign changes in f(x)

We have two sign changes . Keep subtracting by 2 till we get 0

So positive roots are 2 or 0

Remaining 4  are complex. keep subtracting by 2 till we get 0

So complex roots are 4, 2 or 0

To get negative roots, we replace x  with -x in f(x)

To get negative root we look at the sign changes in coefficients and count the sign changes in f(-x)

We have two sign changes . Keep subtracting by 2 till we get 0

So negative roots are 2 or 0

Remaining 4  are complex. keep subtracting by 2 till we get 0

So complex roots are 4, 2 or 0

Answer is  Positive: 4, 2, or 0; negative: 2 or 0; complex: 4, 2, or 0

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